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mestny [16]
4 years ago
9

A cylindrical specimen of a brass alloy having a length of 60 mm (2.36 in.) must elongate only 10.8 mm (0.425 in.) when a tensil

e load of 50,000 N (11,240 lbf ) is applied. Under these circumstances what must be a radius of the specimen
Engineering
1 answer:
Gemiola [76]4 years ago
3 0

The radius of the specimen is 60 mm

<u>Explanation:</u>

Given-

Length, L = 60 mm

Elongated length, l = 10.8 mm

Load, F = 50,000 N

radius, r = ?

We are supposed to calculate the radius of a cylindrical brass specimen in order to produce an elongation of 10.8 mm when a load of 50,000 N is applied. It is necessary to compute the strain corresponding to this

elongation using Equation:

ε = Δl / l₀

ε = 10.8 / 60

ε = 0.18

We know,

σ = F / A

Where A = πr²

According to the stress-strain curve of brass alloy,

σ = 440 MPa

Thus,

sigma = 50,000 / \pi  (r)^2\\\\440 X 10^6 = \frac{50,000}{3.14 X (r)^2}\\\\r = 0.06m\\r = 60mm\\\\\\

Therefore, the radius of the specimen is 60 mm

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